integral of 1/(u^2+u)

Lösung

\int \frac{1}{u ^{2} + u} d u

ln ∣ u ∣ - ln ∣ u + 1 ∣ + C

Schritte zur Lösung

\int \frac{1}{u ^{2} + u} d u

Ermittle den Partialbruch von

\frac{1}{u ^{2} + u} : \frac{1}{u} - \frac{1}{u + 1}

= \int \frac{1}{u} - \frac{1}{u + 1} d u

Wende die Summenregel an:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{1}{u} d u - \int \frac{1}{u + 1} d u

\int \frac{1}{u} d u = ln ∣ u ∣

\int \frac{1}{u + 1} d u = ln ∣ u + 1 ∣

= ln ∣ u ∣ - ln ∣ u + 1 ∣

Füge eine Konstante zur Lösung hinzu

= ln ∣ u ∣ - ln ∣ u + 1 ∣ + C

in v erse l a pl a ce \frac{( s + 2 )}{s ( s + 4 )}

\frac{d}{d x} (tan (5 x + 5))

\int_{0}^{3} \frac{x}{9 - x ^{2}} d x

\int 45 t - \frac{21}{t} d t

\frac{d}{d x} (\frac{2}{x ^{3} + 1})